Problem: $\dfrac{ 5p - 5q }{ 6 } = \dfrac{ -p - 5r }{ -4 }$ Solve for $p$.
Multiply both sides by the left denominator. $\dfrac{ 5p - 5q }{ {6} } = \dfrac{ -p - 5r }{ -4 }$ ${6} \cdot \dfrac{ 5p - 5q }{ {6} } = {6} \cdot \dfrac{ -p - 5r }{ -4 }$ $5p - 5q = {6} \cdot \dfrac { -p - 5r }{ -4 }$ Multiply both sides by the right denominator. $5p - 5q = 6 \cdot \dfrac{ -p - 5r }{ -{4} }$ $-{4} \cdot \left( 5p - 5q \right) = -{4} \cdot 6 \cdot \dfrac{ -p - 5r }{ -{4} }$ $-{4} \cdot \left( 5p - 5q \right) = 6 \cdot \left( -p - 5r \right)$ Distribute both sides $-{4} \cdot \left( 5p - 5q \right) = {6} \cdot \left( -p - 5r \right)$ $-{20}p + {20}q = -{6}p - {30}r$ Combine $p$ terms on the left. $-{20p} + 20q = -{6p} - 30r$ $-{14p} + 20q = -30r$ Move the $q$ term to the right. $-14p + {20q} = -30r$ $-14p = -30r - {20q}$ Isolate $p$ by dividing both sides by its coefficient. $-{14}p = -30r - 20q$ $p = \dfrac{ -30r - 20q }{ -{14} }$ All of these terms are divisible by $2$ Divide by the common factor and swap signs so the denominator isn't negative. $p = \dfrac{ {15}r + {10}q }{ {7} }$